OCR GCSE Physics A: Work done, stored energy and power from forces (J249) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions are aligned to the OCR GCSE (9-1) Gateway Science Physics A specification (J249), Version 5.0, July 2026, first assessment in 2018. They cover statements P2.2m, P2.2n and P2.2o, which are assessed in both tiers. Where a question needs g, use 10 N/kg.

Links: OCR GCSE Physics hub, printable checklist, study guide, revision notes, OCR GCSE Physics forces practice and the free 10-minute diagnostics.

Questions

1. State the equation that links work done, force and distance, and state what one newton-metre is equal to. [2]

2. A student drags a suitcase with a steady horizontal force of 32 N for 14 m along a corridor. Calculate the work done. [2]

3. A tractor pulls a log along the ground. The tractor does 21,000 J of work and the log moves 35 m along the line of the pulling force. Calculate the force. [3]

4. A skier is pulled up a gentle slope by a tow rope. The rope pulls with a force of 120 N along the slope and does 9.0 kJ of work. Calculate the distance the skier moves. [3]

5. A person pushes a loaded trolley across a car park at constant speed. The push is 30 N and the trolley moves 20 m. (a) Calculate the work done by the person. [2] (b) Describe the energy transfer. [2]

6. A builder lifts a bag of cement of mass 8.5 kg vertically through 2.4 m at a steady speed. (a) Calculate the force needed. [1] (b) Calculate the work done. [1] (c) State the gain in gravitational potential energy, and explain why it has this value. [2]

7. A cyclist and bicycle have a total mass of 80 kg and travel at 6.0 m/s. The cyclist brakes and stops. The braking force is a constant 240 N. (a) Calculate the kinetic energy at the start. [2] (b) Calculate the distance travelled while braking. [2]

8. A waiter carries a tray at constant height and constant speed across a restaurant. Explain why the upward force from the waiter’s hand does no work on the tray. [3]

9. Power is the rate at which energy is transferred. (a) Write down the equation for power in terms of work done and time. [1] (b) A hoist lifts a crate of mass 120 kg through 5.0 m in 8.0 s. Calculate the power of the hoist. [4]

10. Two students, A and B, run up the same staircase, which has a vertical height of 4.0 m. Student A has mass 60 kg and takes 12 s. Student B has mass 75 kg and takes 18 s. (a) Calculate the power of each student, to 3 significant figures. [4] (b) Student B says: “I did more work, so I have more power.” Evaluate this statement. [2]

11. A winch pulls a boat of mass 400 kg up a slipway. The pulling force is 1,500 N, along the slope, and the boat moves 9.0 m along the slope in 15 s. The boat rises by a vertical height of 3.0 m. (a) Calculate the work done by the winch. [2] (b) Calculate the gain in gravitational potential energy of the boat. [2] (c) Calculate how much of the work is not stored as gravitational potential energy, and state where it goes. [2] (d) Calculate the power of the winch. [2]

12. A car of mass 900 kg travels at 12 m/s and brakes to a stop in 4.0 s. The brakes provide a constant resistive force of 2,700 N. (a) Calculate the kinetic energy of the car before braking. [2] (b) Calculate the braking distance. [2] (c) Calculate the average power at which energy is transferred while braking. [2]

Answers

1. Work done = force × distance moved along the line of action of the force [1]. 1 N m = 1 J [1] [2] Examiner insight: An answer that says “force × distance” without the phrase about the line of action can still show the idea, but linking the equation to the correct direction of movement makes the answer more secure.

2. W = F × s = 32 × 14 [1]. 448 J [1] [2] Examiner insight: A correct number with no unit, or with N m written where J is expected, is the commonest way to drop a mark on a short calculation.

3. F = W ÷ s [1]. = 21,000 ÷ 35 [1]. 600 N [1] [3] Examiner insight: Rearranging first, then substituting, shows your method clearly even if the arithmetic slips.

4. 9.0 kJ = 9,000 J [1]. s = W ÷ F = 9,000 ÷ 120 [1]. 75 m [1] [3] Examiner insight: Converting kJ to J in your working is the step most often missed; writing it on its own line helps you notice.

5. (a) W = F × s = 30 × 20 [1]. 600 J [1] (b) The trolley moves at constant speed, so its kinetic energy does not change [1]. The 600 J is transferred to thermal energy stores of the wheels, the ground and the surroundings (they warm up) [1] [4] Examiner insight: For “describe the energy transfer”, name the store that gains energy; saying only that energy is “used up” or “lost” does not describe a transfer.

6. (a) At steady speed, force = weight = 8.5 × 10 = 85 N [1] (b) W = 85 × 2.4 = 204 J [1] (c) GPE gain = 204 J [1], because the work done by the lifting force at steady speed is all stored in the gravitational potential energy store (m × g × h = 8.5 × 10 × 2.4 = 204 J) [1] [4] Examiner insight: A statement that the energy is “equal” needs a reason; link the lifting force to the weight rather than just repeating the number.

7. (a) KE = ½ × m × v² = 0.5 × 80 × 6.0² [1]. 1,440 J [1] (b) The braking work removes all the kinetic energy, so s = W ÷ F = 1,440 ÷ 240 [1]. 6.0 m [1] [4] Examiner insight: Squaring only the speed, not the whole product, is the usual slip; write v² as 36 before multiplying.

8. The force from the hand is upwards [1]. The tray moves horizontally, which is at right angles to the force [1]. Nothing moves along the direction of the force, so work done = force × 0 = 0 [1] [3] Examiner insight: Say the movement is at right angles to the force; “nothing is lifted” or “the waiter is not tired” does not explain the physics.

9. (a) Power = work done ÷ time [1] (b) Weight = 120 × 10 = 1,200 N [1]. Work = 1,200 × 5.0 = 6,000 J [1]. P = 6,000 ÷ 8.0 [1]. 750 W [1] [5] Examiner insight: A mass in kg is not a force; turning it into a weight in newtons is the step that links the question to the equation.

10. (a) A: work = 60 × 10 × 4.0 = 2,400 J [1]; P = 2,400 ÷ 12 = 200 W [1]. B: work = 75 × 10 × 4.0 = 3,000 J [1]; P = 3,000 ÷ 18 = 167 W [1] (b) The statement is not correct [1]. Power depends on the work and the time. B did more work but took longer, so A (200 W) has the greater power [1] [6] Examiner insight: An evaluation needs a judgement and a reason supported by your numbers; “B did more work” alone only repeats the claim.

11. (a) W = F × s = 1,500 × 9.0 [1]. 13,500 J [1] (b) GPE = m × g × h = 400 × 10 × 3.0 [1]. 12,000 J [1] (c) 13,500 − 12,000 = 1,500 J [1]. It is transferred to thermal energy stores (friction between boat and slipway, and in the winch) [1] (d) P = W ÷ t = 13,500 ÷ 15 [1]. 900 W [1] [8] Examiner insight: Using the vertical height for GPE and the distance along the slope for work done is the key choice here; mixing them gives a different, wrong comparison.

12. (a) KE = 0.5 × 900 × 12² [1]. 64,800 J [1] (b) s = W ÷ F = 64,800 ÷ 2,700 [1]. 24 m [1] (c) P = W ÷ t = 64,800 ÷ 4.0 [1]. 16,200 W (16.2 kW) [1] [6] Examiner insight: Use the time given in the question, and check that your answer is in watts; mistakes in part (c) often come from dividing by the distance instead of the time.

Where marks are usually lost

• Leaving out units, or using N, N m or W when J is needed.
• Not converting kJ to J before substituting.
• Using mass as a force without multiplying by g.
• Using the slope length where the vertical height is needed, and the other way round.
• Writing that energy is “lost” or “used up” when the answer needs a transfer to thermal stores.
• Treating power as energy, or not linking it to time.
• Not explaining the right angle between force and movement when asked why no work is done.
• Carrying rounded values into later steps and ending with a different third significant figure.

Next steps

• Revise with the revision notes.
• Re-read the study guide.
• Go to the OCR GCSE Physics hub and the printable checklist.
• Try the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR, GCSE (9-1) Gateway Science Physics A (J249) specification, Version 5.0, July 2026, first assessment in 2018. Statements P2.2m, P2.2n and P2.2o.